Question

Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol. 

Answer 1

0.15 M solution of benzoic acid in methanol means, 1000 mL of solution contains 0.15 mol of benzoic acid

 Therefore, 250 mL of solution contains =0.15x250/1000      mol of benzoic acid 

= 0.0375 mol of benzoic acid 

Molar mass of benzoic acid (C₆H₅COOH) = 7 × 12 + 6 × 1 + 2 × 16 = 122 g mol−¹

Hence, required benzoic acid = 0.0375 mol × 122 g mol−¹ = 4.575 g